5 × 1 0 − 5 moles of KMnO 4 in acidic medium is completely consumed to oxidise 5 × 1 0 − 5 moles of X n + into XO 4 − , then what is the value of n ?

Equivalent of KMnO₄ = Equivalent of Xⁿ⁺. 5 × 1 0 − 5 × 5 = 5 × 1 0 − 5 × y , where y is the change in the oxidation number of X. Since y = 5 , then n = 2 .

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$5 \times 1 0^{- 5}$ moles of $KMnO_{4}$ in acidic medium is completely consumed to oxidise $5 \times 1 0^{- 5}$ moles of $X^{n +}$ into $XO_{4}^{-}$ , then what is the value of n ?

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Answer:

Solution:

Equivalent of KMnO₄ = Equivalent of Xⁿ⁺.

$5 \times 1 0^{- 5} \times 5 = 5 \times 1 0^{- 5} \times y$ , where $y$ is the change in the oxidation number of X.

Since $y = 5$ , then $n = 2$ .

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Detection and Estimation of Elements

⏱ 2mℹ️ Source: QB

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